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by knight247 » Sun Oct 16, 2011 5:29 am
How many three digit even numbers N can be formed that are divisible by 9 such that both N and N² have the same units digits?
(A)20
(B)21
(C)22
(D)23
(E)24

OA is A. Detailed explanations would be appreciated.


Another problem can be derived from the above as follows,
How many three digit even integers N can be formed that are divisible by 9 such that both N and N² leave the same remainder when divided by 10?
(A)20
(B)21
(C)22
(D)23
(E)24
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by GmatKiss » Sun Oct 16, 2011 5:50 am
knight247 wrote:How many three digit even numbers N can be formed that are divisible by 9 such that both N and N² have the same units digits?
(A)20
(B)21
(C)22
(D)23
(E)24

OA is A. Detailed explanations would be appreciated.

3 digit numbers divisible by 9,

lets first find out for numbers startting with 1**
108,117,126,135,144,153,162,171,180

so there is a pattern,

8
7
6
5
4
3
2
1
0

and this will be repeated untill 999 (last 3 digit divisible by 9)
in this even numbers will be ending with (0,2,4,6,8) = 100 (total)

N and N^2 has to have same unit digits,
only numbers ending with 0 and 6 satisfies this,
so there is a total of 2*10 = 20 such numbers

IMO: A
Last edited by GmatKiss on Sun Oct 16, 2011 5:52 am, edited 1 time in total.
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by GmatKiss » Sun Oct 16, 2011 5:52 am
Another problem can be derived from the above as follows,
How many three digit even integers N can be formed that are divisible by 9 such that both N and N² leave the same remainder when divided by 10?
(A)20
(B)21
(C)22
(D)23
(E)24

IMO:A as well!! (same reason as 1st problem) please correct me if am wrong!
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by Brent@GMATPrepNow » Sun Oct 16, 2011 6:17 am
knight247 wrote:How many three digit even numbers N can be formed that are divisible by 9 such that both N and N² have the same units digits?
(A)20
(B)21
(C)22
(D)23
(E)24

OA is A. Detailed explanations would be appreciated.


Another problem can be derived from the above as follows,
How many three digit even integers N can be formed that are divisible by 9 such that both N and N² leave the same remainder when divided by 10?
(A)20
(B)21
(C)22
(D)23
(E)24
Both questions here are asking the exact same thing.

First, for an even number, N, to have the same units digit as N^2, then the units digit of N must be either 6 or 0

Next, there are 900 3-digit integers (from 100 to 999).

Of these 900 integers, every 9th integer is divisible by 9. So, there are 100 3-digit integers that are divisible by 9.

Aside: let's list a few of these integers: 108, 117, 126, 135, 144, 153, 162, 171, 180, 189, 198, . . .
Notice that, in this list, each units digit is equally represented.

Now, of the 100 integers that are divisible by 9, we are interested in counting only those that end in a 0 or a 6.

Since there are 10 digits altogether (0,1,2,3,4,5,6,7,8,9), then 2/10 of all 100 integers will end in a 0 or a 6.

So the answer here is [spoiler]20 (A)[/spoiler]

Cheers,
Brent
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by GmatMathPro » Sun Oct 16, 2011 6:53 am
This is not an improvement, but I know you like to see different solutions....


To satisfy "N is even and N and N^2 have the same units digit", N must end in 6 or 0.

To satisfy that N is divisible by 9, the digits must add to a multiple of 9.

for XY0, X+Y=9 or X+Y=18 are possibilities

9:
(X,Y)=(1,8);(2,7)....(9,0) which is 9 ways

18: (X,Y)=(9,9) 1 way

For XY6,
X+Y=12 or X+Y=3

3:
(X,Y)=(1,2);(2,1);(3,0) 3 ways

12:
(X,Y)=(3,9);(4,8).....(9,3) 7 ways

9+1+3+7=

20
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by knight247 » Sun Oct 16, 2011 6:56 am
Yeah Pete...It's definitely not an improvement...I kinda just wrote down all the options like gmatkiss did which is obviously not a viable option during the actual test!
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by GmatMathPro » Sun Oct 16, 2011 8:37 am
I don't know, does it really take that long? I think the point of listing them is just to convince yourself that the multiples of 9 are evenly distributed among the 10 possible units digits. Listing the first 10 or so to establish the pattern shouldn't take more than 30-40 seconds. Once you get that, it's simple to see that it's 2/10 of 100.

Even my way, while admittedly worse, should take less than 2 minutes because it's simple to count the possibilities that go with 6 and 0 at a glance without listing them all.
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by GmatKiss » Sun Oct 16, 2011 9:02 am
Each one has his own method!

My method took me less than 100 seconds.
So eventually, didnt mind looking into any other method.
IMO: It may confuse me.
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